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\(\sqrt{2004}-\sqrt{2003}=\dfrac{1}{\sqrt{2004}+\sqrt{2003}}\)
\(\sqrt{2006}-\sqrt{2005}=\dfrac{1}{\sqrt{2006}+\sqrt{2005}}\)
Mà \(\sqrt{2004}+\sqrt{2003}< \sqrt{2006}< \sqrt{2005}\)
\(\Rightarrow\dfrac{1}{\sqrt{2004}+\sqrt{2003}}>\dfrac{1}{\sqrt{2006}+\sqrt{2005}}\)
\(\Rightarrow\sqrt{2004}-\sqrt{2003}>\sqrt{2006}-\sqrt{2005}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2) \(-x^2+4x-2\)
\(=-\left(x^2-4x+2\right)\)
\(=-\left(x^2-4x+4-2\right)\)
\(=-\left(x-2\right)^2+2\)
Ta có: \(-\left(x-2\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-2\right)^2+2\le2\forall x\)
Dấu "=" xảy ra:
\(\Leftrightarrow-\left(x-2\right)^2+2=2\Leftrightarrow x=2\)
Vậy: GTLN của bt là 2 tại x=2
b) \(\sqrt{2x^2-3}\) (ĐK: \(x\ge\sqrt{\dfrac{3}{2}}\))
Mà: \(\sqrt{2x^2-3}\ge0\forall x\)
Dấu "=" xảy ra:
\(\sqrt{2x^2-3}=0\Leftrightarrow x=\sqrt{\dfrac{3}{2}}=\dfrac{3\sqrt{2}}{2}\)
Vậy GTNN của bt là 0 tại \(x=\dfrac{3\sqrt{2}}{2}\)
...
1:
b: \(4\sqrt{5}=\sqrt{80}\)
\(5\sqrt{3}=\sqrt{75}\)
=>\(4\sqrt{5}>5\sqrt{3}\)
=>\(\sqrt{4\sqrt{5}}>\sqrt{5\sqrt{3}}\)
c: \(3-2\sqrt{5}-1+\sqrt{5}=2-\sqrt{5}< 0\)
=>\(3-2\sqrt{5}< 1-\sqrt{5}\)
d: \(\sqrt{2006}-\sqrt{2005}=\dfrac{1}{\sqrt{2006}+\sqrt{2005}}\)
\(\sqrt{2005}-\sqrt{2004}=\dfrac{1}{\sqrt{2005}+\sqrt{2004}}\)
\(\sqrt{2006}+\sqrt{2005}>\sqrt{2005}+\sqrt{2004}\)
=>\(\dfrac{1}{\sqrt{2006}+\sqrt{2005}}< \dfrac{1}{\sqrt{2005}+\sqrt{2004}}\)
=>\(\sqrt{2006}-\sqrt{2005}< \sqrt{2005}-\sqrt{2004}\)
e: \(\left(\sqrt{2003}+\sqrt{2005}\right)^2=4008+2\cdot\sqrt{2003\cdot2005}=4008+2\cdot\sqrt{2004^2-1}\)
\(\left(2\sqrt{2004}\right)^2=4\cdot2004=4008+2\cdot\sqrt{2004^2}\)
=>\(\left(\sqrt{2003}+\sqrt{2005}\right)^2< \left(2\sqrt{2004}\right)^2\)
=>\(\sqrt{2003}+\sqrt{2005}< 2\sqrt{2004}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Easy
Ta có:
\(\sqrt{2006}-\sqrt{2005}=\frac{2006-2005}{\sqrt{2006}+\sqrt{2005}}=\frac{1}{\sqrt{2006}+\sqrt{2005}}\)
Tương tự cũng có: \(\frac{1}{\sqrt{2007}+\sqrt{2008}}\)
Dễ thấy: \(\sqrt{2005}+\sqrt{2006}< \sqrt{2007}+\sqrt{2008}\)
\(\Rightarrow\frac{1}{\sqrt{2006}+\sqrt{2005}}>\frac{1}{\sqrt{2007}+\sqrt{2008}}\)
Easy
Ta có:
\(\sqrt{2006}-\sqrt{2005}=\frac{2006-2005}{\sqrt{2006}+\sqrt{2005}}=\frac{1}{\sqrt{2006}+\sqrt{2005}}\)
Tương tự cũng có: \(\frac{1}{\sqrt{2007}+\sqrt{2008}}\)
Dễ thấy: \(\sqrt{2005}+\sqrt{2006}< \sqrt{2007}+\sqrt{2008}\)
\(\Rightarrow\frac{1}{\sqrt{2006}+\sqrt{2005}}>\frac{1}{\sqrt{2007}+\sqrt{2008}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\sqrt{2006}-\sqrt{2005}=\frac{1}{\sqrt{2006}+\sqrt{2005}}\)
\(\sqrt{2007}-\sqrt{2006}=\frac{1}{\sqrt{2007}+\sqrt{2006}}\)
Mà : \(\frac{1}{\sqrt{2006}+\sqrt{2005}}>\frac{1}{\sqrt{2007}-\sqrt{2006}}\)
Nến : \(\sqrt{2006}-\sqrt{2005}>\sqrt{2007}-\sqrt{2006}\)
\(\Rightarrow\sqrt{2005}+\sqrt{2007}< 2\sqrt{2006}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\sqrt{2005}+\sqrt{2007}\right)^2=4012+2\sqrt{2005.2007}\)
\(=4012+2\sqrt{\left(2016-1\right)\left(2016+1\right)}=4012+2\sqrt{2016^2-1}\)
\(\left(2\sqrt{2006}\right)^2=4012+4012=4012+2\sqrt{2016^2}\)
=>\(\left(\sqrt{2015}+\sqrt{2017}\right)^2< \left(2\sqrt{2016}\right)^2\Rightarrow\sqrt{2015}+\sqrt{2017}< 2\sqrt{2016}\)
Ta có: \(\sqrt{2006}-\sqrt{2005}=\frac{1}{\sqrt{2006}+\sqrt{2005}}\)
\(\sqrt{2007}-\sqrt{2006}=\frac{1}{\sqrt{2007}+\sqrt{2006}}\)
Mà: \(\frac{1}{\sqrt{2006}+\sqrt{2005}}>\frac{1}{\sqrt{2007}+\sqrt{2006}}\)
Nên: \(\sqrt{2006}-\sqrt{2005}>\sqrt{2007}-\sqrt{2006}\)
=>\(\sqrt{2005}+\sqrt{2007}< 2\sqrt{2006}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng BĐT CAuchy-Schwarz ta có:
Đặt \(A^2=\left(\sqrt{2003}+\sqrt{2005}\right)^2\)
\(\le\left(1+1\right)\left(2003+2005\right)\)
\(=2\cdot4008=8016\)
\(\Rightarrow A^2\le8016\Rightarrow A\le2\sqrt{2004}=B\)
lấy vế đầu trừ vế sau nếu kết quả dương suy ra vế đầu lớn hơn nếu kq âm thì vế sau lớn hơn
có\(\sqrt{2006}-\sqrt{2005}=\frac{\left(\sqrt{2006}-\sqrt{2005}\right)\left(\sqrt{2006}+\sqrt{2005}\right)}{\sqrt{2006}+\sqrt{2005}}\)\(=\frac{1}{\sqrt{2006}+\sqrt{2005}}\)
có\(\sqrt{2005}-\sqrt{2004}=\frac{\left(\sqrt{2005}-\sqrt{2004}\right)\left(\sqrt{2005}+\sqrt{2004}\right)}{\sqrt{2005}+\sqrt{2004}}\)\(=\frac{1}{\sqrt{2005}+\sqrt{2004}}\)
ta lại có 2006>2005\(\Rightarrow\sqrt{2006}>\sqrt{2005}\)có 2005>2004\(\Rightarrow\sqrt{2005}>\sqrt{2004}\)
\(\Rightarrow\sqrt{2006}+\sqrt{2005}>\sqrt{2005}+\sqrt{2004}\)\(\Rightarrow\frac{1}{\sqrt{2006}+\sqrt{2005}}< \frac{1}{\sqrt{2005}+\sqrt{2004}}\)
\(\Rightarrow\sqrt{2006}-\sqrt{2005}>\sqrt{2005}-\sqrt{2004}\)